【DP】Decode Ways

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A message containing letters from A-Z is being encoded to numbers using the following mapping:

'A' -> 1'B' -> 2...'Z' -> 26

Given an encoded message containing digits, determine the total number of ways to decode it.

For example,
Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).

The number of ways decoding "12" is 2.

解法一：DFS，超时

public class Solution {        private int count = 0;        public boolean isDigit(String s){        if(s==null) return false;        int len = s.length();        if(len==0 || len>2) return false;                int sum = 0;        for(int i=0; i<len; i++){            sum = sum * 10 + s.charAt(i) - '0';        }        if(sum >26 || sum ==0) return false;        return true;    }        public void dfs(String s, int pos){        if(pos == s.length()){            count++;            return;        }                for(int i=1; i<3 && pos+i<s.length(); i++){            String str = s.substring(pos, pos+i);            if(isDigit(str)){                dfs(s, pos+i);            }        }    }        public int numDecodings(String s) {        dfs(s, 0);        return count;    }}

解法二：DP

public class Solution {        public boolean isValid(String s){        if(s.charAt(0) == '0') return false;                int sum = 0;        for(int i=0; i<s.length(); i++){            sum = sum * 10 + s.charAt(i) - '0';        }        if(sum <= 26) return true;        return false;    }        public int numDecodings(String s) {        if(s == null ) return 0;        int count = 0;        int len = s.length();        if(len == 0) return 0;                int d[] = new int[len+1];        d[0] = 1;                if(isValid(s.substring(0, 1)))            d[1] = 1;        else d[1] = 0;                for(int i=2; i<=len; i++){            if(isValid(s.substring(i-1, i)))            d[i] += d[i-1];            if(isValid(s.substring(i-2, i))){                d[i] += d[i-2];            }        }                return d[len];    }}

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